Question

Handwritten Please

A 0.045-kg bullet traveling at a speed of 456 m/s is fired vertically into a sponge (mass of sponge = 0.06 kg) that is initial at rest. Assuming that the bullet is embedded in the sponge, how high would the combined bullet and sponge rise after the perfectly inelastic collision?

Answer 1

here,

mass of bullet, mb = 0.045 kg

velocity of bullet, vb = 456 m/s

mass of sponge, ms = 0.06 kg

From conservation of momentum we have :
before collision = after collision
mb*vb + ms*vs = (mb + ms) * V

velocity of system, V = mb*vb/(mb + ms)

velocity of system, V = 0.045*456/(0.045 + 0.06)

velocity of system, V = 195.43 m/s

From conseravtion of energy :

kinetic energy of system = gain in PE

0.5 * (mb + ms) * v^2 = (mb + ms) * g * h

height, h = 0.5*v^2/g

height, h = 0.5 * 195.43^2/9.81

height, h = 1946.63 m